Đáp án:
$x=-\dfrac{5\pi}{12}+k\pi$ $(k\in\mathbb Z)$
Giải thích các bước giải:
$(\sin2x+\sqrt3\cos2x)^2-\cos\left({2x-\dfrac{\pi}6}\right)=5$
$\Leftrightarrow (\sin2x+\sqrt3\cos2x)^2-\dfrac{\sqrt3}2\cos 2x-\dfrac12\sin2x=5$
$\Leftrightarrow (\sin2x+\sqrt3\cos2x)^2-\dfrac12(\sqrt3\cos2x+\sin2x)=5$
$\Leftrightarrow\left[\begin{array}{I}\sin2x+\sqrt3\cos2x=2,5\text{(do }1^2+\sqrt3^2=4<2,5^2\text{ (loại))}\\\sin2x+\sqrt3\cos2x=-2\text{(do }1^2+\sqrt3^2=4\ge(-2)^2\text{ (thỏa mãn)) (1)}\end{array}\right.$
(1) tương đương:
$\dfrac12\sin2x+\dfrac{\sqrt3}2\cos 2x=-1$
$\Leftrightarrow \sin2x\cos\dfrac{\pi}3+\sin\dfrac{\pi}3\cos2x=-1$
$\Leftrightarrow\sin\left({2x+\dfrac{\pi}3}\right)=-1$
$\Leftrightarrow 2x+\dfrac{\pi}3=-\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=-\dfrac{5\pi}{12}+k\pi$ $(k\in\mathbb Z)$.
